# About : $\int_{0}^{\infty}\frac{1}{x^{2}+\ln\left(x\right)^{2}}dx>\int_{1}^{\infty}\left(\frac{1}{x^{2}+\ln\left(x\right)^{2}}\right)dx+1$

Prove that :

$$\int_{0}^{\infty}\left(\frac{1}{x^{2}+\ln\left(x\right)^{2}}\right)dx>\left(\int_{1}^{\infty}\left(\frac{1}{x^{2}+\ln\left(x\right)^{2}}\right)dx\right)+1$$

This inequality is quite nice for the eyes but I don't think we can find an exact result for the two integrals (I don't ask for that).

My attempt :

I have tried to use the inequality for $$x\geq1$$ :

$$\ln{x}\leq(x-1)\sqrt[3]{\frac{2}{x^2+x}}.$$

A proof of it can be found here Which is greater $\frac{13}{32}$ or $\ln \left(\frac{3}{2}\right)$ and is due to user Michael Rozenberg .

Unfortunately the difficulty is the same and I'm stuck here .Perhaps we can use the substitution $$y=\ln(x)$$ .

Question : How to show the first inequality ?

• Maybe these related questions will help. Aug 8, 2021 at 11:36
• So you are asking if $\int_0^1\frac{1}{x^{2}+\ln\left(x\right)^{2}}\,dx > 1$ ? Aug 8, 2021 at 18:26
• @MartinR I think a solution only using a calculator is preferred. A calculator can calculate values of elementary functions such as $\sin \sqrt{2}$ (no special functions such as the Lambert W function). Aug 11, 2021 at 14:19
• The question is also equivalent to $$\int_0^{ + \infty } {\frac{1}{{1 + x^2 }}\frac{{dx}}{{1 + W_0 (x)}}} > 1$$ where $W_0$ is the principal branch of the Lamber $W$-function.
– Gary
Aug 19, 2021 at 7:08

We need to prove that $$\int_0^1 \frac{1}{x^2 + \ln^2 x}\, \mathrm{d} x > 1.$$
Fact 1: For all $$y \in (0, \mathrm{e}]$$, $$\frac{-y^4 + 64y^3 - 318y^2 + 400y + 35}{132y + 48} \ge y(1 - \ln y)^2.$$ (Note: LHS is the (4, 1)-Pade approximant of $$y(1 - \ln y)^2$$ at $$y = 1$$.)
With the substitution $$y = \mathrm{e}x$$, using Fact 1, we have \begin{align*} & \int_0^1 \frac{1}{x^2 + \ln^2 x}\, \mathrm{d} x\\ =\, & \int_0^\mathrm{e} \frac{\mathrm{e}}{y^2 + \mathrm{e}^2 (1 - \ln y)^2}\, \mathrm{d} y\\ \ge\, & \int_0^\mathrm{e} \frac{132\mathrm{e}y^2 + 48\mathrm{e}y}{(132 - \mathrm{e}^2)y^4 + (64\mathrm{e}^2 + 48)y^3 - 318\mathrm{e}^2 y^2 + 400\mathrm{e}^2y + 35\mathrm{e}^2}\, \mathrm{d} y\\ \ge\, & \int_0^\mathrm{e} \frac{132\mathrm{e}y^2 + 48\mathrm{e}y}{(132 - \mathrm{e}^2)(y + 1/12)(y + 463/64)[(y - 69/44)^2 + (55/54)^2]}\, \mathrm{d} y \tag{1} \\[8pt] =\, & \frac{1253273472\mathrm{e}}{911936989\mathrm{e}^2 - 120375682548} \int_0^\mathrm{e} \frac{1}{12y + 1}\, \mathrm{d} y\\[8pt] &\quad + \frac{29129459463143424\mathrm{e}}{38952637956653\mathrm{e}^2 - 5141748210278196} \int_0^\mathrm{e} \frac{1}{64y + 463}\, \mathrm{d} y \\[8pt] &\quad + \frac{372594816\mathrm{e}}{2487336612548656836 - 18843459185974673\mathrm{e}^2}\\[8pt] &\quad\quad \times \int_0^\mathrm{e} \frac{842186377401732y + 746090344086823}{1411344y^2 - 4426488y + 4934869}\, \mathrm{d} y \tag{2}\\ >\, & 1 \end{align*} where in (1) we have used (easy): for all $$y\in [0, \mathrm{e}]$$, \begin{align*} &(132 - \mathrm{e}^2)y^4 + (64\mathrm{e}^2 + 48)y^3 - 318\mathrm{e}^2 y^2 + 400\mathrm{e}^2y + 35\mathrm{e}^2\\ \le\, & (132 - \mathrm{e}^2)(y + 1/12)(y + 463/64)[(y - 69/44)^2 + (55/54)^2]. \end{align*} Note: The three integrals in (2) admit closed form expressions in terms of $$\ln$$ and $$\arctan$$ only.